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\begin{document}

\title{Math 243, Lecture 12}
\maketitle

 \section{ Probability distributions} 

To be systematic about probabilities in the continuous setting, we have to 
formalize the notion of distribution.  

\begin{definition}
\begin{enumerate}
\item A distribution for a continuous random variable can be
any non-negative function $D$ where the total area under the function is $1$.  
\item If $X$ is distributed according to a distribution $D$ then
$P (a \leq X \leq b)$ is the area between the line $x = a$, the line $x = b$,
the graph of $D$, and the $x$-axis.  
\item The sample space for such a variable is all $x$ where $D$ is non-zero.  
\item If $D$ is equal to either $0$ or one other fixed number, we say $D$
is a uniform distribution.  
\end{enumerate}
\end{definition}

%\includegraphics{Figures/uniform0-3.eps}  

\begin{example}
Revisit the example about numbers between 0 and 3. 
\begin{itemize}
\item Our sample space    is the closed interval $[0,3]$.  
\item Our probability distribution    is the function 
\[
p (t) = \left\{ \begin{array}{ll}
1/3 & 0 \leq t \leq 3\\
0  & \mbox{else}
\end{array} \right.  
\]
\end{itemize}
Again, what is the probability that some randomly
chosen number is between 1.5 and 2.4?   Etc.  
\end{example}


\begin{example}
Which of the following are distributions? 
\begin{itemize}
\item $f(x) = 1$ for $x$ between $-2$ and $-1$, and zero otherwise.  
\item $f(x) = 1$ over the whole real number line.  
\item $f(x) = \frac{x}{2}$ for $x$ between 0 and 2 and zero otherwise.   
\item $f(x) = \frac{x}{2}$ for $x$ between $-1$ and $1$ and zero otherwise.  
\end{itemize}
\end{example}

\begin{example}
Suppose that $X$ is distributed according to 
$$D = \begin{cases} \frac{1}{12} \;\; 1 \leq x \leq 3 \\
\frac{1}{2} \;\; 3 \leq x \leq 4 \\  \frac{1}{9} \;\; 4 \leq x \leq 7 \\
0 \;\; {\text{otherwise}}
\end{cases}$$
\begin{itemize}
\item What is the state space?  
\item What is $P(2 \leq X \leq 3)$?  
\item Which is more likely, that $X$ is between $3$ and $4$ or that $X$
is between $5$ and $7$?   
\end{itemize}
\end{example}

Compare the answer to the last question to the answer you would get if
$X$ were uniformly distributed.  

One way to get a distribution is to take the histogram of a data set and divide
by the total number of individuals so that the area is one.   
The probabilities computed correspond to percentages of the data.  

\begin{example}
Translate between probability and percentile questions about 
scores of students taking an exam.  
\end{example}



 \section{ Normal distributions}  

\begin{example}{[Main Example]}
$\mathbf{Z}$ is a number chosen with normal distribution $N (\mu ,\sigma)$.  
We'll use $N (0,1)$ as our first example.  
%\includegraphics{Figures/normal0-1.eps}
\begin{itemize}
\item Our random event is the choice of a number.  
\item Our random variable $\mathbf{Z}$ is the value of the number.  
\item Our sample space is all possible real numbers.  
\item Our probability distribution is the function 
\[
e^{-t^{2}/2}/\sqrt{2\pi }.  
\]
\item What is $P (\mathbf{Z}>3)$? (Hint: use 68-95-99.7 rule.)   % = .0015$ by 
\item What is $P (0 \leq Z \leq 2)$?    % can be read off of Table A with a bit of work.
\item $P (\mathbf{Z}>100)$?   
%The probability  $P
%(\mathbf{Z}>22)$ is less that $\frac{1}{10^{99}}$, and will come out
%as $0$ on our calculators.
\end{itemize}
\end{example}

 

Doing calculations with normal distributions works much as before.  
Instead of calculating percentiles, we are calculating probabilities.  


\begin{example}
Let our random event be the random choice of a U.S. adult male,
assuming a probability distribution which is 
 approximately normal $N (70,4).$.  
\begin{enumerate}
% \includegraphics{Figures/normal70-4.pdf}
\item What is our random variable?    Our state space?  
(theoretical vs. actual)  
\item What is the probability that our randomly chosen man
is between 70 inches and 73 inches tall?  
\end{enumerate}
\end{example}


 \section{ Sampling distributions}  

A basic practice in statistics is to take a distribution and 
{\em sample} from it.  

\begin{definition}
Given a population,  the  {sampling
distribution} for samples of size $n$ is the probability distribution
of some parameter 
as we take values $n$ times.  
\end{definition} 

We go thoroughly through an example looking
at sampling distributions and comment
thoroughly as we go along.  

Consider the collection of numbers 
\[
\mathbf{S} = \{2,2,5,6,7 \}.
\]
 What is the mean?    Ans: 4.4.    
 
 We can consider all samples from our collection of size $1$.  There
are 5 of them and again, the mean of the samples is again $4.4$.  The
standard deviation is 2.302. 


We can now consider all samples from $\mathbf{S}$ of size $2,$
and take the mean of each sample.    Our samples are
\begin{multline*} 
\{\{2,2 \},\{2,5 \},\{2,6 \},\{2,7 \},\{2,5 \},\{2,6 \}, \\
 \{2,7 \},\{5,6 \},\{5,7 \},\{6,7 \} \}.
\end{multline*}   
Our means are 
\[
\{2,3.5, 4, 4.5, 3.5, 4, 4.5, 5.5, 6, 6.5 \}
\]  
Notice that the spread is smaller, even though there are more numbers. 
The mean is still 4.4, and the standard deviation is now 1.329.  

The probability distribution of these numbers is an example of
 {\em sampling distribution}.     In this case 
it is for the mean of a sample of 2 from the set $\mathbf{S}$.   
 There are 7 possible outcomes which \emph{don't} have equal probabilities.  

\noindent \begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$X$ &  2 & 3.5 & 4 & 4.5 & 5.5 & 6 & 6.5   \\ \hline
Probability & .1  & .2 &  .2 & .2 & .1 & .1 & .1\\ \hline  
\end{tabular}
 

Now consider all samples of size 3, and their means.    There are
again 10 such samples (you might recall this from our digression 
on binomial coefficients), which we won't list.     The means for these 10
samples are 
\[
\{3, 3.333, 3.667, 4.333, 4.667, 4.333, 4.667, 5, 5, 6 \}
\]  
The standard deviation is now only .886.  

Our  {sampling distribution} for samples of $3$ out of our original
group $S$ is 

\noindent \begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$X$ &  3 & 3.333 & 3.667 & 4.333 & 4.667 & 5 & 6   \\ \hline
Probability & .1  & .1 &  .1 & .2 & .2 & .2 & .1\\ \hline
\end{tabular}

 

\begin{example}
Calculate the sampling distribution for the mean of samples of two 
out of the data set $\{-1, 0, 0, 2, 3, 6\}$.  
\end{example}

 

The process of taking a sampling distribution is not an easy one to understand the
first time you see it; go through this example and the book carefully.   

There are two important cases of sampling distributions.   The first is one we
have seen in a different language.  

\begin{theorem}
The sampling distribution measuring number of heads in $n$ coin flips is called
the {\em binomial distribution}, with values giving by Pascal's triangle.  
As $n$ gets large, this sampling distribution approaches a normal distribution.  
\end{theorem}

We won't elaborate on this example - this is just recasting our coin flipping
observations in a new language.  The next example is a new one for us. 

\begin{theorem}
The sampling distribution for the mean of $n$ measurements of some data which
is known to be following a distribution of $N( \mu, \sigma)$ is itself normal, given
by $N(\mu, \sigma / \sqrt{n})$.  
\end{theorem}

\begin{example}
For men's heights, distributed according to $N (70,4)$, what is the probability that 
nine men chosen at random have an average height between $70$ and $73$?  
\end{example}

Compare this with our previous answer as to the probability of finding a single
man with such height.


\end{document}
 